Integral of 2sinx

The integral of 2sinx is equal to -2cosx + C. This integral can be evaluated using the formula for the integral of sin x. We know that integration is the reverse of the process of differentiation and hence, the integral of 2sinx is also called the antiderivative of 2sinx. We know that the integral of sinx is equal to -cosx + C. Therefore, the integration of 2sinx is given by -2cosx + C, where C is the constant of integration.

In this article, we will compute the integral of 2sinx and derive its formula. We will also evaluate the definite integration of 2sinx with different limits and go through a few solved examples based on a similar concept for a better understanding.

The integral of 2sinx can be calculated using the formula for integral of sinx which is given by, ∫sin x dx = -cos x + C, where C is the integration constant. Now, to find the integration of 2sinx, we use this formula and mathematically, write the integral of 2sinx as ∫2sinx dx = -2cosx + C, with C as the constant of integration. Let us now explore the formula for the same in the next section.

As we know that ∫sin x dx = -cos x + C, therefore the formula for the integral of 2sinx is written as ∫2sinx dx = -2cosx + C, where ∫ symbol denotes the integration of 2sinx, dx denotes that the variable is x and C is the integration constant. The image below gives the integral of the 2sinx formula:

Integral of 2sinx Proof

Now that we know the integral of 2sinx is equal to -2cosx + C, next in this section, we will prove this result using different formulas of integration. We will use the following antiderivative rules and formulas:

• ∫sinx dx = -cosx + C
• ∫kf(x) dx = k ∫f(x) dx

Using the above formulas, we have

∫2sinx dx = 2 ∫sinx dx

= 2[-cosx + K]

= -2cosx + 2K

= -2cosx + C, where C = 2K is the integration constant.

Hence, we have proved that the integral of 2sinx is equal to -2cosx + C.

To verify that the integration of 2sinx is -2cosx + C, we can take the derivate of -2cosx + C. We know that the derivative of cos x is equal to -sin x, that is, d(cos x)/dx = -sin x and the derivative of constant function is equal to zero. Therefore, we have d(-2cosx + C)/dx = d(-2cosx)/dx + dC/dx = -2d(cosx)/dx + 0 = -2(-sinx) = 2sinx. Hence, verified.

We know that the formula for the integration of 2sinx is equal to -2cosx + C. Now, we will determine the definite integral of 2sinx with limits from 0 to pi. To find this definite integral, we will substitute the limits into the formula of the integral of 2sinx and take their difference.

\(\begin{align}\int_{0}^{\pi}2\sin x \ dx &= \left [ -2 \cos x + C\right ]_{0}^{\pi}\\&=(-2\cos \pi+C)-(-2\cos 0 + C)\\&=-2\times (-1)+C+2\times1-C\\&=2+2\\&=4 \end{align}\)

Hence, the value of the definite integral of 2sinx with limits from 0 to π is equal to 4.

Important Notes on Integral of 2sinx

• The integral of 2sinx is mathematically written as ∫2sinx dx = -2cosx + C.
• We can find the integration of 2sinx using the integral of sinx.

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What is the Integral of 2sinx?

The integral of 2sinx is equal to -2cosx + C, where C is the integration constant. This integral can be evaluated using the formula for the integral of sin x.

What is the Formula for the Integration of 2sinx?

The integral of 2sinx can be calculated using the formula for integral of sinx which is given by, ∫2sin x dx = -2cos x + C, where C is the integration constant.

How To Find Integral of 2sinx?

We can find the integral of 2sinx by using the following antiderivative rules and formulas:

• ∫sinx dx = -cosx + C
• ∫kf(x) dx = k ∫f(x) dx

Using the above formulas, we have the integral of 2sinx to be equal to -2cosx + C, with C as the integration constant.

What is the Indefinite Integral of 2sinx With Respect to cosx?

The indefinite integral of 2sinx with respect to cosx is determined as ∫2sinx d(cosx) = ∫2sinx (-sinx) dx = ∫-2sin²x dx = - ∫(1 - cos2x) dx = -(x - sin2x/2) + C = -x + (1/2)sin2x + C. Hence, the integral of 2sinx with respect to cosx is given by -x + (1/2)sin2x + C, where C is the constant of integration.

What is the Value of Integral of 2sinx From 0 to pi?

The value of the definite integral of 2sinx with limits from 0 to π is equal to 4. This integral value is determined by substituting the limits into the formula of the integral of 2sinx and subtracting the two.